A few administrative notes before we review the day’s material: I will not be holding office hours this Wednesday. And there are no classes next Monday, when my usual set of office hours are. But I’ve decided to do a sort of experiment: I don’t plan on reviewing for the exam specifically next week, but a large portion of the class has said that they would come to office hours on Monday if I were to have them. So I’m going to hold them to that – I’ll be in Kassar House 105 (the MRC room) from 7-8:30 (or so, later perhaps if there are a lot of questions), and this will dually function as my office hours and a sort of review session.

But this comes with a few strings attached: firstly, I’ll be willing to answer any question, but I’m not going to prepare a review; secondly, if there is poor turnout, then this won’t happen again. Alrighty!

The rest is after the fold –

The topic of the day was differentiation! The three questions of the day were –

  1. Differentiate the following functions:
    1. e^x
    2. e^{e^x}
    3. e^{e^{e^x}}
    4. \sin x
    5. \sin (\sin x)
    6. \sin (\sin (\sin x))
  2. A particle moves along a line with its position described by the function s(t) = a_0t^2 + a_1t + a_2. If we know that it’s acceleration is always 20 m/s/s, that its velocity at t = 1 is -10 m/s, and its position at t = 2 is 20 m. What are a_0, a_1, a_2?
  3. Given that u(x) = (x^2 + x + 2, what are the following:
    1. \frac{d}{dx} (u(x))^2
    2. \frac{d}{dx} (u(x))^n
    3. \frac{d}{dx} (5 + x^3)^{-3}
    4. \frac{d}{dx} ((u(x))^n)^m

Question 1

This is all about the chain rule. Please note that this is a big deal, so if you have any trouble at all with the chain rule, seek extra help. The derivative of e^x is e^x. To compute the derivative of e^{e^x}, we might think of u(x) = e^x, so that we have e^u. The derivative of e^u will be e^u u', which gives us e^{e^x}e^x. Let’s look at the other way of understanding the chain rule to compute the derivative of e^{e^{e^x}}. The “outer function” is e^{(\cdot)}. It’s derivative is just itself. The first “inner function” is e^{e^x}. We have just computed its derivative above (it’s e^{e^x} e^x). So we multiply them together to get e^{e^{e^x}}e^{e^x}e^x.

Similarly, the derivative of \sin x is $\cos x$. The derivative of $\sin \sin x$ requires the chain rule. On the one hand, the outer function is \sin, and the derivative of \sin is $\cos$. So we know we will have a \cos (\sin x) in the answer. The inner function is also \sin x, so we need to multiply by its derivative. The final answer will be \cos (\sin x )\cos x. To compute the derivative of \sin \sin \sin x, we again use the chain rule. I will again use helper functions, to illustrate their use. We might call u(x) = \sin \sin x, so that we are computing the derivative of \sin (u). Then we get \cos u u'. We happen to have computed u' just a moment ago, so the final answer is \cos \sin \sin x \cos \sin x \sin x.

Question 2

The key idea of this question is to remember that the function s(t) gives position at time t. So its derivative gives a result in terms of position per time, the velocity. And the derivative of velocity will give a result in terms of position per time per time, or acceleration. So the velocity of our particle is 2a_0t + a_1, and the acceleration is 2a_0. Since we know that the acceleration is always 20, we know that 2a_0 = 20 so that a_0 = 10. The velocity at t = 1 is -10, so we know that 2(10)(1) + a_1 = -10, so that a_1 = -30. Finally, our position at time t = 2 is 20, so that 4(10) + 2(-30) + a_2 = 20, so that a_2 = 40. I used different numbers between the two classes, so don’t pay too much attention if the exact details are different between one class and the other.

Question 3

This is more about the chain-rule! This is sort of an explicit example of helper functions. We first want to compute the derivative of u(x)^2. By the chain rule, this will be 2u(x)u'(x). What is u'(x)?. It’s 2x + 1. So the derivative of u(x)^2 is 2(x^2 + x + 2)(2x + 1). This is a single case of the slightly more general u(x)^n. Here, the power rule tells us that the derivative will be nu(x)^{n-1}u'(x), which is n (x^2 + x + 2)^{n-1}(2x + 1).

The idea behind the third question is to see if we can work out the same sort of idea, but without starting with a helper function. (It’s perfectly fine to always use helper functions to use the chain rule – that’s not a problem at all). The derivative of (5 + x^3)^{-3} will be -3(5 + x^3)^{-4}(3x^2). If we want to see the use of helper functions, call v(x) = 5 + x^3, so that we are computing the derivative of v^{-3}. The derivative will be -3v^{-4}v', which is exactly what we have above.

I look forward to seeing some of you on Monday, and happy studying!